Sets: Class 11 Maths · Chapter 1

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

105 min
56 practice questions
26 MCQs

What you will learn in Sets

Decide whether a collection is well defined, and describe a set in roster form and in set-builder form.
Recognise empty, singleton, finite, infinite and equal sets, and count the distinct elements of a finite set.
Tell membership (∈) apart from inclusion (⊂), list every subset of a small set and spot proper subsets.
Name the standard subsets of R and convert between interval notation and set-builder form.
Choose a suitable universal set and draw Venn diagrams for sets and the relations between them.
Find unions, intersections, differences and complements, and apply their laws, including the distributive and De Morgan laws.
Prove set identities by showing each side is a subset of the other, and disprove false claims with a counterexample.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 1, pp. 1–23. Not affiliated with or endorsed by NCERT.

1.1 Introduction

A selector picking a squad, a railway app that shows only the trains leaving Bhopal after 6 pm, a phone that files contacts under "Family" and "School": each one forms a collection and then settles, object by object, who is in and who is out. Mathematics turns this everyday habit into a precise tool called a set (समुच्चय).

Sets are the shared language of the rest of this book: a function (Chapter 2) is a set of ordered pairs, the solutions of an inequality (Chapter 5) form a set on the number line, and an event in probability (Chapter 14) is a set of outcomes. This chapter builds the vocabulary for describing, comparing and combining sets. The theory began with Georg Cantor (1845–1918) in the 1870s; his story is in the historical note.

1.2 Sets and their Representations

You group things all the time without thinking about it: the songs in a playlist, the passengers on a local train, the books on your desk. A mathematical collection, however, has to pass one strict test. Pick any object at all, and the question "does it belong?" must have a definite yes or no answer.

Test a few descriptions. "The states of India that share a border with Nepal" passes: Uttarakhand, Uttar Pradesh, Bihar, West Bengal and Sikkim belong, and Kerala does not. "The roots of $x^{2} - 7 x + 12 = 0$ " passes too, with exactly 3 and 4 belonging. "The greatest spinners India has produced" fails, because two cricket fans can honestly disagree. "The tall students in your class" fails for the same reason, yet it is easy to repair: "the students in your class taller than 165 cm" is a perfectly good set.

Definition

Set

A set is a collection of objects that is well defined: for every object, it is clear whether or not it belongs to the collection. The objects are called the elements or members of the set.

ExampleThe prime numbers less than 20 form a set; the "interesting" numbers less than 20 do not.

We name sets with capital letters such as $A$ , $B$ , $S$ , $X$ and elements with small letters such as $a$ , $p$ , $x$ . The symbol $\in$ (a stylised Greek epsilon) means "is an element of" or "belongs to", and $\notin$ means "is not an element of". If $P$ is the set of primes less than 20, then $7 \in P$ but $9 \notin P$ , since $9 = 3 \times 3$ . If $M$ is the set of months with exactly 30 days, then June $\in M$ and July $\notin M$ .

Table 1.1Number sets used throughout this bookNCERT 1.2 · p. 2

Symbol	The set of	Some members	Not a member
$𝐍$	natural numbers	$1, 2, 3, 100$	$0$
$𝐙$	integers	$- 3, 0, 5$	$\frac{1}{2}$
$𝐐$	rational numbers	$- \frac{2}{3}, 0.25, 7$	$\sqrt{2}$
$𝐑$	real numbers	$\sqrt{2}, π, - 4.1$	$\sqrt{- 1}$
$𝐙^{+}$	positive integers	$1, 2, 3$	$0$
$𝐐^{+}$	positive rationals	$\frac{3}{4}, 5$	$- \frac{3}{4}$
$𝐑^{+}$	positive reals	$π, 0.001$	$0$

In this book the natural numbers start at 1, so $0 \notin 𝐍$ and $𝐙^{+}$ is the same set as $𝐍$ . You will meet numbers outside $𝐑$ , such as $\sqrt{- 1}$ , in Chapter 4.

There are two standard ways to describe a set. In roster form we write out the elements one by one between curly braces, with commas in between: the positive divisors of 28 form the set ${1, 2, 4, 7, 14, 28} .$ Three conventions keep the notation honest.

Order is irrelevant. ${1, 2, 4, 7, 14, 28}$ and ${28, 7, 1, 14, 4, 2}$ describe the same set.
Each element is written once. The letters of the word COMMITTEE form the set ${C, O, M, I, T, E},$ not a list of nine letters.
Three dots continue a clear pattern. The positive multiples of 5 are ${5, 10, 15, 20, \dots},$ and the dots promise that the pattern never stops.

Set-builder form (समुच्चय निर्माण रूप) lists nothing. Instead it gives a test: a condition that the members of the set pass and every other object fails. The divisors of 28 become ${x : x \in 𝐍 and x divides 28};$ in words, "every natural number $x$ that divides 28". The letter $x$ is only a placeholder that any other letter could replace, and some books use a vertical bar for the colon, as in ${x ∣ x > 0}$ .

3/4Described by a rule that all six numbers satisfy and no other number does.

Fig. 1.1The same set in words, as a list and as a rule.

Six number tiles, 1, 2, 4, 7, 14 and 28, under the heading "the positive divisors of 28". In the first step the tiles are scattered, a collection described only in words. In the second step they line up in a row between two large curly braces, which is roster form. In the third step the row stays and the set-builder rule, x such that x is a natural number that divides 28, appears below it. In the fourth step the tiles swap places to the order 14, 1, 28, 2, 7, 4 and a dashed second copy of 7 appears below the first one with the note "written once": the set has not changed.

Each form has its job. Roster form suits small sets and lets you check membership at a glance. Set-builder form handles sets that are large, infinite or impossible to list, such as ${x : x \in 𝐑, 0 < x < 1},$ the real numbers between 0 and 1. Moving between the two forms is a basic skill, and the examples below practise both directions.

Example 1NCERT Example 1 · p. 3

Roots of a quadratic as a set

Write the solution set of $2 x^{2} - 5 x - 3 = 0$ in roster form.

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 3

From a rule to a list

Write ${n : n \in 𝐙, - 2 \leq 3 n + 1 < 11}$ in roster form.

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 4

Spotting the rule

Write $A = {2, 5, 10, 17, 26, \dots}$ in set-builder form.

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 4

A pattern in fractions

Write ${\frac{3}{4}, \frac{5}{9}, \frac{7}{16}, \frac{9}{25}, \frac{11}{36}}$ in set-builder form.

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 4

Matching two descriptions

Match each set in roster form with the same set in set-builder form.

Roster form	Set-builder form
(i) ${K, O, L, A, T}$	(a) ${x : x \in 𝐙, x^{2} = 4}$
(ii) ${1}$	(b) letters of the word KOLKATA
(iii) ${1, 3, 9}$	(c) ${x : x \in 𝐍, x divides 9}$
(iv) ${- 2, 2}$	(d) ${x : x \in 𝐑, x^{3} = 1}$

Try it yourself first. Then reveal the solution one step at a time.

1.3 The Empty Set

Some perfectly clear descriptions select nothing at all. Take ${x : x \in 𝐍 and x + 5 = 2} :$ the only number with $x + 5 = 2$ is $- 3$ , which is not natural, so no element qualifies. Or take the Class 11 students of your school who were born on 30 February. The description is precise, and the answer is "nobody".

Definition

Empty set

A set with no elements at all is called the empty set; the names null set and void set are also used. We write it $ϕ$ or ${}$ .

Example ${x : x \in 𝐑, x^{2} + 1 = 0} = ϕ,$ because the square of a real number is never $- 1$ .

Three more empty sets, each with its reason:

${x : x \in 𝐍, x < 1} = ϕ,$ since the smallest natural number is 1.
${x : x \in 𝐐, x^{2} = 3} = ϕ,$ since the only solutions, $\pm \sqrt{3}$ , are irrational.
The set of primes that are multiples of 4 is $ϕ$ : every multiple of 4 is divisible by 2 and larger than 2, so it is not prime.

Be careful with braces. $ϕ$ is an empty box, ${0}$ is a box holding the number 0, and ${ϕ}$ is a box holding an empty box. Only the first is empty (Fig. 1.2).

3/3 ${ϕ}$ has one element, the empty set itself.

Fig. 1.2Only

ϕ

is empty;

{0}

and

{ϕ}

each have one element.

Three rounded boxes in a row. In the first step only the left box shows: it is empty, labelled phi equals empty braces, with the note "no element". In the second step a middle box appears holding a large 0, labelled set containing 0, with the note "one element". In the third step a right box appears holding a smaller empty box, labelled set containing phi, also with the note "one element".

Quick check

1 of 2

Which of these sets are empty?

{x : x \in 𝐍, 2 x = 7}

{x : x \in 𝐙, x^{2} = 0}

{x : x \in 𝐑, x^{2} < 0}

{ϕ}

Select all that apply

Show answer

Answer

(A) ${x : x \in 𝐍, 2 x = 7}$
(C) ${x : x \in 𝐑, x^{2} < 0}$

2 x = 7

needs

x = 3.5

, which is not natural, and no real square is negative. But

x^{2} = 0

has the solution 0, and

{ϕ}

has one element, the empty set.

1.4 Finite and Infinite Sets

How big is a set? For ${a, e, i, o, u}$ the answer is 5. We write $n (A)$ for the number of distinct elements of a set $A$ , so a repeated entry is never counted twice. If $L$ is the set of letters of BANANA, then $L = {B, A, N}$ and $n (L) = 3.$

Some sets are huge yet still countable in principle. Kerala has 14 districts, and the grains of sand on Juhu beach run into unimaginable numbers, yet that count stops somewhere. Other sets never run out: however far you list the multiples of 7, there is always a next one (Fig. 1.3).

Definition

Finite and infinite sets

A set is finite if it is empty or if counting its elements ends at some natural number $n$ ; then $n (A) = n,$ and $n (ϕ) = 0.$ A set that is not finite is infinite.

ExampleThe days of the week form a finite set with $n = 7$ ; the positive multiples of 7 form an infinite set.

3/3The days stop at $n = 7$ , a finite set. The multiples of 7 have no last element, an infinite set.

Fig. 1.3A finite set runs out of new elements; an infinite set never does.

Two rows of tiles. In the first step each row shows three tiles: Mon, Tue, Wed in the top row and 7, 14, 21 in the bottom row. In the second step both rows fill up: the top row reaches Sun after seven tiles, and the bottom row continues to 49. In the third step the top row gets the label n = 7, while the bottom row ends in three dots and an arrow running off the edge, labelled "no last element".

More examples: the districts of Kerala form a finite set with $n = 14$ ; the solutions of $x^{3} = x$ form ${- 1, 0, 1},$ finite with $n = 3$ ; the points on a circle and the lines through a given point form infinite sets.

Roster form cannot list an infinite set in full, so we write enough elements to make the pattern unmistakable and add dots: ${7, 14, 21, \dots}$ for the positive multiples of 7, or ${\dots, - 6, - 3, 0, 3, 6, \dots}$ for the multiples of 3 among the integers. Some infinite sets resist even this. The real numbers between 0 and 1 follow no pattern that dots could continue, and Cantor proved that no list whatsoever can contain them all (see the historical note). For such sets, set-builder form or interval notation (Section 1.6.2) is the only way.

Example 6NCERT Example 6 · p. 7

Finite or infinite?

Decide whether each set is finite or infinite.

${x : x \in 𝐍, (x - 2) (x + 5) = 0}$
${x : x \in 𝐍, x^{2} = 5}$
${x : x \in 𝐙, ∣ x ∣ < 4}$
${x : x \in 𝐍, x is a multiple of 3}$
${x : x \in 𝐑, 0 < x < 0.001}$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Which set is infinite?

the lines through the point

(1, 2)

the people living in India todaythe real roots of

x^{4} = 16

the letters of the English alphabet

Show answer

Answer (A) the lines through the point $(1, 2)$

Through one point there is a line in every direction, so there are infinitely many. India's population is enormous but finite, and

x^{4} = 16

has only the real roots

\pm 2

1.5 Equal Sets

When do two descriptions give the same set? Only the elements matter: not the wording, not the order, not the repeats.

Definition

Equal sets

Two sets are equal, $A = B$ , when belonging to one always means belonging to the other: each element of $A$ lies in $B$ and each element of $B$ lies in $A$ . If even one element breaks this, $A \neq B$ .

Example ${x : x \in 𝐙, x^{2} = 4}$ and ${2, - 2}$ are equal, although one is given by an equation and the other by a list.

Equality ignores everything except membership. The letters of NAINITAL and the letters of LATIN give the same set, ${N, A, I, T, L},$ although one word has eight letters and the other five (Fig. 1.4). Listing an element a second time adds nothing new.

3/3Each letter of one set is in the other and vice versa, so the sets are equal.

Fig. 1.4Different words, same set of letters.

Two rows of letter tiles: N, A, I, N, I, T, A, L on top and L, A, T, I, N below. In the first step all the tiles are shown. In the second step the second N, the second I and the second A of NAINITAL fade, leaving five distinct letters. In the third step a line joins each remaining top letter to the same letter in the bottom row: every letter has exactly one partner, so the two sets are equal.

To show $A \neq B$ , one witness is enough: an element that lies in one set and not the other. Equal sets always have the same number of elements, but the converse fails: ${1, 2}$ and ${3, 4}$ both have two elements and share none.

Example 7NCERT Example 7 · p. 8

Picking out equal sets

Which of these sets are equal?

$P = {- 2, 2}$
$Q = {x : x \in 𝐙, x^{2} = 4}$
$R = {x : x \in 𝐑, x^{3} = 8}$
$S$ = the set of even primes
$T = {x : x \in 𝐙, ∣ x ∣ \leq 2}$

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 8

Equal or not?

Decide whether the sets in each pair are equal, and justify your answer.

$X$ , the set of letters of TEEN, and $Y$ , the set of letters of NET.
$A$ , the set of prime factors of 45, and $B = {x : x \in 𝐍, x^{2} - 8 x + 15 = 0} .$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

n (A) = n (B),

then

A = B

TrueFalse

Show answer

Answer False

{1, 2}

and

{3, 4}

have two elements each but are not equal. The reverse statement is true: equal sets have the same number of elements.

1.6 Subsets

Every square is a rectangle, but not every rectangle is a square. Every prime greater than 2 is odd, but 9 is odd and not prime. Statements like these, one set lying entirely inside another, are what the idea of a subset captures.

Definition

Subset

A set $A$ is a subset of a set $B$ , written $A \subset B$ , if each member of $A$ belongs to $B$ as well. If at least one member of $A$ lies outside $B$ , then $A$ is not a subset of $B$ , written $A \subset̸ B$ .

Example ${2, 4} \subset {1, 2, 3, 4},$ but ${2, 5} \subset̸ {1, 2, 3, 4},$ because 5 is missing.

The arrow $\Rightarrow$ means "implies", so the definition can be written compactly: $A \subset B$ if $x \in A \Rightarrow x \in B .$ Notice the asymmetry. To prove $A \subset B$ you must account for every element of $A$ , but to prove $A \subset̸ B$ a single element of $A$ outside $B$ is enough.

Inclusion in both directions forces equality: if $A \subset B$ and $B \subset A$ , neither set can hold an element the other lacks. Conversely, equal sets are subsets of each other. A two-way statement like this uses the double arrow $\Leftrightarrow$ , spoken "if and only if":

\begin{array}{c} A = B \\ ⇕ \\ A \subset B and B \subset A \end{array}

This two-way check is the standard way to prove that two sets are equal, and you will use it in Miscellaneous Examples.

Two facts follow straight from the definition. Every set is a subset of itself, $A \subset A$ , since each element of $A$ is in $A$ . The empty set is a subset of every set, $ϕ \subset A$ : to break the rule, $ϕ$ would need an element outside $A$ , and it has no elements at all.

When $A \subset B$ but $A \neq B$ , the set $B$ has at least one element that $A$ lacks. Then $A$ is a proper subset (उचित उपसमुच्चय) of $B$ , and $B$ is a superset of $A$ . A set with exactly one element, such as ${7}$ , is a singleton.

Some examples. $𝐙 \subset 𝐐$ , since every integer $m$ equals $\frac{m}{1}$ . The squares form a proper subset of the rectangles, and ${4}$ is a proper subset of ${4, 9}$ . For ${1, 2}$ and ${2, 3}$ , neither set is a subset of the other.

Example 9NCERT Example 9 · p. 10

Inside or not?

Let $P = {2, 6}$ , $Q = {2, 4, 8}$ and $R = {2, 4, 6, 8, 10} .$ Write $\subset$ or $\subset̸$ between the sets in each pair: (i) $Q, R$ (ii) $R, Q$ (iii) $ϕ, P$ (iv) $P, Q$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10NCERT Example 10 · p. 10

Two sets, neither inside the other

Let $A$ be the set of natural-number multiples of 4 and $B$ the set of natural-number multiples of 6. Is $A \subset B$ ? Is $B \subset A$ ?

Try it yourself first. Then reveal the solution one step at a time.

The symbols $\in$ and $\subset$ are easy to confuse, and the difference matters once sets have other sets as elements. Let $A = {2, {2, 3}, 5} .$ This set has three elements: the number 2, the set ${2, 3}$ and the number 5 (Fig. 1.5). So ${2, 3} \in A$ . But ${2, 3} \subset̸ A,$ because its element 3 is not one of the three elements of $A$ : the 3 sits inside the inner set, not in $A$ itself. On the other hand ${2, 5} \subset A$ , and so is ${{2, 3}}$ , the set whose only element is ${2, 3}$ .

3/4Is each element of ${2, 3}$ an element of $A$ ? 2 is, but 3 is not, so ${2, 3} \subset̸ A .$

Fig. 1.5Being an element of

A

and being a subset of

A

are different tests.

A large rounded bag labelled A holds three things: a loose 2, a smaller inner bag containing 2 and 3, and a loose 5. In the first step the bag is shown with A written in set notation below it. In the second step the inner bag is highlighted: the set of 2 and 3 is one element of A. In the third step two tiles on the right test whether that set is a subset of A: the tile 2 is marked "in A" and the loose 2 is ringed, but the tile 3 is marked "not in A", because 3 appears only inside the inner bag. In the fourth step the tiles 2 and 5 are both marked "in A" and both loose numbers are ringed, so the set of 2 and 5 is a subset of A.

Example 11NCERT Example 11 · p. 10

Membership passes along; inclusion need not

Suppose $A \in B$ and $B \subset C$ . (a) Show that $A \in C$ . (b) Show by an example that $A \subset C$ need not hold.

Try it yourself first. Then reveal the solution one step at a time.

1.6.1 Subsets of set of real numbers

The number systems you know fit inside one another. In set notation:

$𝐍 = {1, 2, 3, \dots},$ the natural numbers;
$𝐙 = {\dots, - 2, - 1, 0, 1, 2, \dots},$ the integers;
$𝐐 = {x : x = \frac{p}{q}, p, q \in 𝐙, q \neq 0},$ the rational numbers;
$𝐓 = {x : x \in 𝐑, x \notin 𝐐},$ the irrational numbers.

Rational numbers come in many disguises: $0.75 = \frac{3}{4},$ $0.333 \dots = \frac{1}{3},$ $\sqrt{9} = \frac{3}{1}$ and $- 1.2 = - \frac{6}{5}$ all belong to $𝐐$ . A real number that is not rational is irrational; $\sqrt{3}$ , $\sqrt{7}$ , $π$ and $0.1010010001 \dots$ are examples, and their decimals never settle into a repeating block. The basic relations are $𝐍 \subset 𝐙 \subset 𝐐 \subset 𝐑,$ $𝐓 \subset 𝐑$ and $𝐍 \subset̸ 𝐓$ (Fig. 1.6).

4/4 $𝐑$ is $𝐐$ together with the irrationals $𝐓$ , such as $\sqrt{2}$ and $π$ .

Fig. 1.6Each number system contains the one before it;

𝐐

and

𝐓

together make up

𝐑

Nested rounded rectangles. In the first step only the innermost box N is shown, holding 1, 2 and 9. In the second step a larger box Z appears around it and adds 0 and minus 5. In the third step a still larger box Q surrounds Z and adds minus 2.5 and three quarters. In the fourth step the outer box R appears together with a dashed box T beside Q, holding root 2, root 7 and pi; the boxes Q and T do not overlap, and together they fill R.

1.6.2 Intervals as subsets of R

Many useful subsets of $𝐑$ are unbroken stretches of the number line, and they have a compact notation. Let $a$ and $b$ be real numbers with $a < b$ .

Definition

Intervals

The open interval $(a, b) = {x : x \in 𝐑, a < x < b}$ leaves out both end points. The closed interval $[a, b] = {x : x \in 𝐑, a \leq x \leq b}$ keeps both. The half-open intervals $[a, b) = {x : a \leq x < b}$ and $(a, b] = {x : a < x \leq b}$ keep exactly one end point.

Example $[2, 5)$ contains 2 and 4.99 but not 5.

A square bracket means "included" and a round bracket "excluded". On a number line we draw a filled dot for an included end point and a hollow dot for an excluded one (Fig. 1.7). Each of these intervals contains infinitely many real numbers, and the number $b - a$ is called its length, whichever brackets it has.

3/4Half-open: $- 1 \leq x < 3;$ the filled end belongs, the hollow end does not.

Fig. 1.7Filled dot: included. Hollow dot: excluded.

A number line marked from minus 3 to 5 with a thick highlighted segment from minus 1 to 3. In the first step both end dots are hollow: the open interval from minus 1 to 3. In the second step both dots are filled: the closed interval. In the third step the left dot is filled and the right dot hollow: the interval that includes minus 1 but not 3. In each of these three steps the length 4 is noted below the line. In the fourth step the right dot disappears and the highlight runs past 5 with an arrowhead: the ray of all numbers from minus 1 onwards, which includes minus 1.

Intervals can also run on forever in one direction: $(2, \infty)$ is the set of reals greater than 2, $(- \infty, 5]$ the set of reals that are at most 5, and $(- \infty, \infty)$ is the whole of $𝐑$ . The symbol $\infty$ is not a real number, so it always takes a round bracket; $[2, \infty]$ has no meaning.

Brackets make subset questions quick. For instance, $(- 1, 4] \subset [- 2, 6),$ since every $x$ with $- 1 < x \leq 4$ also satisfies $- 2 \leq x < 6.$ Converting between notations is just as quick: ${x : x \in 𝐑, - 2 \leq x < 3} = [- 2, 3),$ and $(0, 7] = {x : x \in 𝐑, 0 < x \leq 7} .$

Example

An interval from two conditions

Write ${x : x \in 𝐑, x \geq - 1 and 2 x - 3 < 5}$ as an interval, and find its length.

Try it yourself first. Then reveal the solution one step at a time.

1.7 Universal Set

Every discussion has a background. In a chapter on quadratic equations, every solution set you write down lives inside $𝐑$ ; in a survey of how a school's Class 11 students travel, every set of interest sits inside the set of those students.

Definition

Universal set

A universal set, written $U$ , is a set that contains all the sets being considered in a given context, so that each of them is a subset of $U$ .

ExampleFor a study of the multiples of 7 below 100, $U$ could be ${1, 2, \dots, 99}$ or the whole of $𝐍$ .

$U$ is a choice made to suit the context, and the choice can change answers: with $U = 𝐑$ the set ${x : x^{2} + 1 = 0}$ is empty, but Chapter 4 enlarges the number system so that this equation has two solutions.

1.8 Venn Diagrams

Relations between sets are easier to see than to read. In 1880 the English logician John Venn (1834–1923) introduced a simple picture for them, now called a Venn diagram (वेन आरेख). The whole rectangle is the universal set; each subset is drawn as a circle (any closed loop would do) inside it, and elements can be written in the regions where they belong.

A circle drawn completely inside another shows a subset, overlapping circles show sets that share elements, and circles drawn apart show sets with nothing in common. In Fig. 1.8 the universal set is $U = {1, 2, \dots, 12},$ $A$ is the set of multiples of 3 in $U$ and $B$ the set of multiples of 6. Every multiple of 6 is a multiple of 3, so the circle for $B$ sits inside the circle for $A$ : $B \subset A$ .

3/3 $B = {6, 12}$ lies inside $A$ , showing $B \subset A$ .

Fig. 1.8A circle inside a circle shows a subset.

A rectangle labelled U contains the numbers 1 to 12. In the first step there are no circles, only the twelve numbers. In the second step a circle labelled A is drawn around 3, 6, 9 and 12, the multiples of 3. In the third step a smaller circle labelled B appears inside A, around 6 and 12, the multiples of 6. The other eight numbers stay outside both circles.

You will draw many more Venn diagrams in the next section, where each operation on sets becomes a shaded region.

Quick check

1 of 2

In a Venn diagram, the circle for

P

lies completely inside the circle for

Q

. What does this show?

Q \subset P

P \subset Q

P \in Q

P

and

Q

have no common element

Show answer

Answer (B) $P \subset Q$

Every element of

P

is drawn inside

Q

, which is exactly

P \subset Q

. Drawing one circle inside another shows inclusion, not membership:

P \in Q

would make

P

a single element of

Q

1.9 Operations on Sets

In arithmetic, an operation takes two numbers and returns a new number, as addition turns 7 and 8 into 15. Set operations work the same way: they take two sets and return a set. From now on every set is a subset of some universal set $U$ , and in a Venn diagram the result of each operation is a shaded region.

1.9.1 Union of sets

Suppose the school choir and the debate team are both travelling to an inter-school festival by one bus. The bus must carry everyone who is in the choir, in the debate team, or in both, and a student who is in both still needs only one seat. That combined list is the union (सम्मिलन) of the two sets.

Example 12NCERT Example 12 · p. 14

Listing a union

Let $A = {3, 6, 9, 12, 15}$ and $B = {5, 10, 15} .$ Find $A \cup B$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 13NCERT Example 13 · p. 14

Union with a subset

Let $A$ be the set of positive divisors of 12 and $B$ the set of positive divisors of 4. Show that $A \cup B = A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 14

Who boards the bus?

Four Class 11 students sing in the school choir, $C$ = {Anjali, Farhan, Gurpreet, Meera}, and three are in the debate team, $D$ = {Farhan, Lakshmi, Meera}. Find $C \cup D$ and say what it represents.

Try it yourself first. Then reveal the solution one step at a time.

Definition

Union

The union of sets $A$ and $B$ , written $A \cup B$ and read " $A$ union $B$ ", is the set of elements that belong to at least one of $A$ and $B$ : $A \cup B = {x : x \in A or x \in B} .$

Example ${1, 2, 3} \cup {3, 4} = {1, 2, 3, 4};$ the common element 3 is written once.

The word "or" here is inclusive: an element in both sets counts. In a Venn diagram, $A \cup B$ is the whole region covered by the two circles (Fig. 1.9).

3/3Shade only the overlap: $A \cap B = {1, 2, 3, 6} .$

Fig. 1.9Union: everything in either circle. Intersection: only the overlap.

A rectangle U holds two overlapping circles, A on the left and B on the right. The overlap holds 1, 2, 3 and 6; the rest of A holds 4 and 12; the rest of B holds 9 and 18; the numbers 5 and 7 lie outside both circles. In the first step nothing is shaded. In the second step both circles are shaded completely: the union, with eight numbers. In the third step only the overlap is shaded: the intersection, with the four numbers 1, 2, 3 and 6.

Union obeys five laws. Each one follows from the meaning of "or", and a quick Venn sketch makes each one obvious.

Commutative law: $A \cup B = B \cup A,$ since "in $A$ or in $B$ " says the same as "in $B$ or in $A$ ".
Associative law: $(A \cup B) \cup C = A \cup (B \cup C);$ both are the elements in at least one of the three sets, so we simply write $A \cup B \cup C$ .
Identity law: $A \cup ϕ = A$ . The empty set adds nothing, so $ϕ$ is the identity for $\cup$ , as 0 is for $+$ .
Idempotent law: $A \cup A = A$ ; combining a set with itself brings in no new element.
Law of $U$ : $U \cup A = U$ , because $A$ already lies inside $U$ .

1.9.2 Intersection of sets

Back to the festival. The organisers also need the students who are in the choir and in the debate team, because those students cannot be in two halls at once. That common part is the intersection (सर्वनिष्ठ).

Example 15NCERT Example 15 · p. 15

A common element

Find $A \cap B$ for $A = {3, 6, 9, 12, 15}$ and $B = {5, 10, 15},$ the two sets used in Example 12.

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 15

In both teams

Use the choir and debate-team sets $C$ and $D$ from Example 14 to find $C \cap D$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 15

Intersection with a subset

Let $A = {x : x \in 𝐍, x \leq 20}$ and let $B$ be the set of perfect squares in $A$ . Find $A \cap B$ and show that it equals $B$ .

Try it yourself first. Then reveal the solution one step at a time.

Definition

Intersection

The intersection of $A$ and $B$ , written $A \cap B$ and read " $A$ intersection $B$ ", collects the elements that $A$ and $B$ have in common: $A \cap B = {x : x \in A and x \in B} .$

Example ${1, 2, 3} \cap {2, 3, 4} = {2, 3} .$

In a Venn diagram, $A \cap B$ is the region where the circles overlap (the last step of Fig. 1.9). Two sets may have no element in common, like the students of Section A and of Section B of Class 11, or like $𝐐$ and $𝐓$ . Such sets are called disjoint (असंयुक्त), and their circles are drawn apart.

Definition

Disjoint sets

Sets $A$ and $B$ are disjoint if they have no common element, that is, if $A \cap B = ϕ$ .

Example ${10, 20, 30}$ and ${15, 25, 35}$ are disjoint.

Intersection has its own laws.

Commutative law: $A \cap B = B \cap A .$
Associative law: $(A \cap B) \cap C = A \cap (B \cap C);$ both are the elements common to all three sets.
Laws of $ϕ$ and $U$ : $ϕ \cap A = ϕ$ and $U \cap A = A$ . Here $U$ plays the part of the identity, as 1 does for $\times$ .
Idempotent law: $A \cap A = A$ .
Distributive law: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C);$ intersection distributes over union.

The distributive law looks like $a (b + c) = a b + a c$ in arithmetic, and Fig. 1.10 builds both sides step by step to show that they cover the same region. Unlike arithmetic, the law still holds with the operations swapped: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C),$ so union also distributes over intersection, while $a + b c \neq (a + b) (a + c)$ for most numbers.

5/5Their union, $(A \cap B) \cup (A \cap C),$ covers exactly the region of the left side.

Fig. 1.10Both sides of the distributive law shade the same region.

Three overlapping circles A, B and C inside a rectangle U. In the first step nothing is shaded. In the second step all of B and C is shaded: the set B union C. In the third step only the part of A that lies in B or in C is shaded: a curved region made of the overlap of A with B and the overlap of A with C. In the fourth step the overlap of A and B is hatched with lines sloping one way and the overlap of A and C with lines sloping the other way; the small centre where all three circles meet carries both hatchings. In the fifth step the solid shading of the third step returns under the hatching and covers exactly the hatched area.

InteractiveVenn lab

A Venn diagram with two or three circles inside a rectangle, a keypad with the letters A, B, C and U, the empty set, the union, intersection, minus and complement signs and brackets, and two expression boxes. The region of each expression is shaded and described in words, with the elements it contains for sample sets inside U = {1, 2, …, 20}. When both boxes hold an expression, a panel says whether they cover the same regions for every choice of sets, or names a region where they differ. A challenge mode shows an expression, asks you to tap the regions it covers, and checks your answer.

Fig. 1.11Type any expression in

A

B

and

C

to see its region; compare two expressions to test a law, or shade a region yourself and check it.

1.9.3 Difference of sets

Some questions ask what one set has and the other lacks. Which choir members do not debate? Which primes are not odd?

Definition

Difference of sets

The difference $A - B$ , read " $A$ minus $B$ ", is the set of elements of $A$ that are not in $B$ : $A - B = {x : x \in A and x \notin B} .$

Example ${1, 2, 3, 4} - {3, 4, 5} = {1, 2} .$

Example 18NCERT Example 18 · p. 16

Primes that are not odd

Let $A$ be the set of primes less than 15 and $B$ the set of odd natural numbers less than 15. Find $A - B$ and $B - A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 19NCERT Example 19 · p. 17

Letters of two cities

Let $X$ be the set of letters of INDORE and $Y$ the set of letters of DELHI. Find $X - Y$ and $Y - X$ .

Try it yourself first. Then reveal the solution one step at a time.

So the order matters: in general $A - B \neq B - A .$ In a Venn diagram, $A - B$ is the part of circle $A$ outside circle $B$ (Fig. 1.12).

The two differences and the intersection cut $A \cup B$ into three pieces. Every element of $A \cup B$ is in $A$ only, in $B$ only, or in both, and never in two of these at once. So $A - B$ , $A \cap B$ and $B - A$ are pairwise disjoint, and together they make up $A \cup B$ .

4/4 $X - Y$ , $X \cap Y$ and $Y - X$ never overlap, and together they fill $X \cup Y$ .

Fig. 1.12The two differences and the intersection are three non-overlapping pieces of the union.

A rectangle holds overlapping circles X and Y. The overlap holds I, D and E; the rest of X holds N, O and R; the rest of Y holds L and H; the letters K and M lie outside both. In the first step nothing is shaded. In the second step the left crescent, X minus Y, is shaded. In the third step the right crescent, Y minus X, is shaded instead. In the fourth step the left crescent, the overlap and the right crescent are shaded in three different tints: three separate pieces that together fill both circles.

1.10 Complement of a Set

A school library is open from Monday to Friday. Take $U$ to be the set of the seven days of the week and $A$ the set of days on which the library is open. The days it is closed are exactly the days of $U$ that are not in $A$ : Saturday and Sunday. This leftover part of the universal set is the complement (पूरक) of $A$ .

Definition

Complement

For a subset $A$ of the universal set $U$ , the complement $A^{'}$ collects everything in $U$ that $A$ leaves out: $A^{'} = {x : x \in U, x \notin A},$ which is the difference $U - A$ .

ExampleWith $U$ the days of the week and $A$ = {Monday, …, Friday}, $A^{'}$ = {Saturday, Sunday}.

The complement depends on $U$ . If $U$ were only the six days from Monday to Saturday, the same $A$ would have $A^{'}$ = {Saturday}. In a Venn diagram, $A^{'}$ is everything inside the rectangle and outside the circle (Fig. 1.13).

2/3 $A^{'}$ = {Sat, Sun}: the part of $U$ outside $A$ .

Fig. 1.13The complement is the part of

U

outside the set.

A rectangle U holds the seven days of the week. A circle labelled A contains Mon, Tue, Wed, Thu and Fri; Sat and Sun lie outside it. In the first step nothing is shaded. In the second step everything inside the rectangle but outside the circle is shaded: the complement of A, holding Sat and Sun. In the third step the shading moves back inside the circle, showing that the complement of the complement is A again.

Example 20NCERT Example 20 · p. 19

Finding a complement

Let $U = {1, 2, \dots, 12}$ and let $A$ be the set of multiples of 3 in $U$ . Find $A^{'}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 21NCERT Example 21 · p. 19

A complement in words

Let $U$ be the set of Class 11 students of a school and $A$ the set of those who come to school by the school bus. Describe $A^{'}$ .

Try it yourself first. Then reveal the solution one step at a time.

Now take the complement of $A^{'}$ in Example 20: the elements of $U$ outside ${1, 2, 4, 5, 7, 8, 10, 11}$ are 3, 6, 9 and 12, which is $A$ again. This always happens. Removing the outside of the outside gives back the inside, so $(A^{'})^{'} = A$ for every subset $A$ of $U$ . Note also that $A^{'}$ is itself a subset of $U$ .

Example 22NCERT Example 22 · p. 19

Complement of a union

Let $U = {a, b, c, d, e, f, g},$ $A = {a, b, c}$ and $B = {c, d, e} .$ Find $A^{'}$ , $B^{'}$ , $A^{'} \cap B^{'}$ and $(A \cup B)^{'},$ and compare the last two.

Try it yourself first. Then reveal the solution one step at a time.

Example 22 is no accident. An element lies outside $A \cup B$ exactly when it lies outside $A$ and outside $B$ . A matching statement holds for intersections, and the pair is named after the British mathematician Augustus De Morgan (1806–1871), who was born in Madurai.

LawDe Morgan's laws

For any subsets $A$ and $B$ of $U$ , the complement of $A \cup B$ is $A^{'} \cap B^{'}$ , and the complement of $A \cap B$ is $A^{'} \cup B^{'}$ . Taking complements swaps union and intersection.

We prove the first law by following an arbitrary element $x$ of $U$ . Saying $x \in (A \cup B)^{'}$ means $x \notin A \cup B$ , that is, $x$ lies in neither set: $x \notin A$ and $x \notin B$ . That is the same as $x \in A^{'}$ and $x \in B^{'}$ , which says $x \in A^{'} \cap B^{'} .$

Read forwards, the chain shows $(A \cup B)^{'} \subset A^{'} \cap B^{'};$ read backwards, it shows $A^{'} \cap B^{'} \subset (A \cup B)^{'} .$ The second law follows in the same way, because $x \notin A \cap B$ means $x \notin A$ or $x \notin B$ .

A Venn diagram makes the first law visible (Fig. 1.14): shading the outside of both circles gives the same region as overlapping the outside of $A$ with the outside of $B$ .

5/5 $(A \cup B)^{'}$ and $A^{'} \cap B^{'}$ shade exactly the same region.

Fig. 1.14

(A \cup B)^{'}

and

A^{'} \cap B^{'}

are the same region.

Two overlapping circles A and B in a rectangle U. In the first step both circles are shaded: the union. In the second step only the region outside both circles is shaded: the complement of the union. In the third step everything outside circle A is hatched with lines sloping one way: the complement of A. In the fourth step everything outside circle B is also hatched, with lines sloping the other way; only the region outside both circles carries both hatchings, which is the intersection of the two complements. In the fifth step the solid shading of the second step returns under the faded hatching and fills exactly the cross-hatched region.

The complement obeys four laws, each of which a quick Venn sketch confirms.

Complement laws: $A \cup A^{'} = U$ and $A \cap A^{'} = ϕ;$ a set and its complement fill $U$ without overlapping.
De Morgan's laws: the complement of $A \cup B$ is $A^{'} \cap B^{'}$ , and the complement of $A \cap B$ is $A^{'} \cup B^{'}$ ; taking complements swaps $\cup$ and $\cap$ .
Double complement: $(A^{'})^{'} = A;$ the outside of the outside is the inside.
Laws of $ϕ$ and $U$ : $ϕ^{'} = U$ and $U^{'} = ϕ$ ; leaving out nothing gives all of $U$ , and nothing lies outside $U$ .

Practice · 7 questions

Practice 1.5: questions with answers

1
Short answerEasyNCERT Exercise 1.5, Q1
Let $U = {1, 2, \dots, 10},$ $A = {2, 4, 6, 8, 10},$ $B = {1, 2, 3, 4, 5}$ and $C = {3, 6, 9} .$ Find each set.
1. $A^{'}$
  Show solution
  Answer ${1, 3, 5, 7, 9}$
  The odd numbers of $U$ .
2. $(A \cup C)^{'}$
  Show solution
  Answer ${1, 5, 7}$
  $A \cup C = {2, 3, 4, 6, 8, 9, 10} .$
3. $(A \cap B)^{'}$
  Show solution
  Answer ${1, 3, 5, 6, 7, 8, 9, 10}$
  $A \cap B = {2, 4} .$
4. $(B - C)^{'}$
  Show solution
  Answer ${3, 6, 7, 8, 9, 10}$
  $B - C = {1, 2, 4, 5} .$
Short answerEasyNCERT Exercise 1.5, Q2
Let $U = {p, q, r, s, t, u, v, w} .$ Find the complements of (a) ${p, q}$ (b) ${q, s, u, w}$ (c) ${r, s, t, u, v}$ (d) $U$ .
Show solution
(a) ${r, s, t, u, v, w}$ (b) ${p, r, t, v}$ (c) ${p, q, w}$ (d) $ϕ$ , since nothing in $U$ lies outside $U$ .
3
Short answerMediumNCERT Exercise 1.5, Q3
Taking $𝐍$ as the universal set, write the complement of each set.
1. the natural multiples of 4
  Show solution
  Answerthe natural numbers not divisible by 4: ${1, 2, 3, 5, 6, 7, 9, \dots}$
  Remove 4, 8, 12, … from $𝐍$ .
2. ${x : x \in 𝐍, x > 10}$
  Show solution
  Answer ${1, 2, \dots, 10}$
  What is left are the natural numbers up to 10.
3. ${x : x \in 𝐍, 3 x - 2 = 13}$
  Show solution
  Answerall natural numbers except 5
  The set is ${5}$ , since $3 x = 15$ .
4. the natural numbers divisible by both 2 and 3
  Show solution
  Answerthe natural numbers not divisible by 6
  Divisible by 2 and by 3 means divisible by 6.
4
ProofMediumNCERT Exercise 1.5, Q4
Let $U = {1, 2, \dots, 10},$ $A = {1, 3, 5, 7, 9}$ and $B = {1, 2, 5, 10} .$ Verify each of De Morgan's laws.
1. $(A \cup B)^{'} = A^{'} \cap B^{'}$
  Show solution
  AnswerBoth sides equal ${4, 6, 8}$ .
  $A \cup B = {1, 2, 3, 5, 7, 9, 10},$ so $(A \cup B)^{'} = {4, 6, 8} .$
  $A^{'} = {2, 4, 6, 8, 10}$ and $B^{'} = {3, 4, 6, 7, 8, 9},$ so $A^{'} \cap B^{'} = {4, 6, 8} .$
2. $(A \cap B)^{'} = A^{'} \cup B^{'}$
  Show solution
  AnswerBoth sides equal ${2, 3, 4, 6, 7, 8, 9, 10} .$
  $A \cap B = {1, 5},$ so $(A \cap B)^{'}$ is everything else in $U$ .
  $A^{'} \cup B^{'} = {2, 4, 6, 8, 10} \cup {3, 4, 6, 7, 8, 9},$ which is the same eight numbers.
5
DiagramMediumNCERT Exercise 1.5, Q5
Draw two overlapping circles $A$ and $B$ inside a rectangle $U$ , and describe the region for each set. Check your answers in the Venn lab.
1. $A \cap B^{'}$
  Show solution
  AnswerThe crescent of $A$ outside $B$ , the same region as $A - B$ .
  Inside $A$ and outside $B$ .
2. $(A - B)^{'}$
  Show solution
  AnswerEverything except that crescent: all of $B$ together with the region outside both circles.
  The complement of the crescent is the rest of the rectangle.
3. $A^{'} \cup B$
  Show solution
  AnswerThe same region as in (b).
  Outside $A$ , or inside $B$ : that covers everything except the part of $A$ outside $B$ . Indeed $(A - B)^{'} = (A \cap B^{'})^{'} = A^{'} \cup B$ by De Morgan's law.
4. $(A \cap B) \cup (A \cup B)^{'}$
  Show solution
  AnswerThe overlap of the circles together with the region outside both.
  These are the elements in both sets or in neither.
Short answerMediumNCERT Exercise 1.5, Q6
Let $U$ be the set of all quadrilaterals in a plane, and $A$ the set of quadrilaterals with at least one angle different from $90^{\circ}$ . What is $A^{'}$ ?
Show solution
Answer $A^{'}$ is the set of rectangles (squares included).
A quadrilateral is outside $A$ exactly when none of its angles differs from $90^{\circ}$ , that is, when all four angles are right angles. Those quadrilaterals are the rectangles, and squares are special rectangles.
7
Quick answerEasyNCERT Exercise 1.5, Q7
Fill in the blanks to make true statements about a subset $A$ of $U$ .
1. $U^{'} \cup A =$
  Show answer
  Answer A
  $U^{'} = ϕ$ , and $ϕ \cup A = A$ .
2. $ϕ^{'} \cup A =$
  Show answer
  Answer U
  $ϕ^{'} = U$ , and $U \cup A = U$ .
3. $(A^{'})^{'} \cap A^{'} =$
  Show answer
  Answer φ
  $(A^{'})^{'} = A,$ and $A \cap A^{'} = ϕ .$
4. $A \cup (A^{'} \cap U) =$
  Show answer
  Answer U
  $A^{'} \cap U = A^{'},$ and $A \cup A^{'} = U .$

Miscellaneous Examples

Example 23NCERT Example 23 · p. 20

Same letters, different words

The words ASSAM and MASS look quite different. Prove that both are spelt with exactly the same letters, that is, that their sets of letters are equal.

Try it yourself first. Then reveal the solution one step at a time.

Example 24NCERT Example 24 · p. 21

Listing every subset

List all the subsets of $A = {x : x \in 𝐙, x^{3} = 4 x} .$

Try it yourself first. Then reveal the solution one step at a time.

4/4 $2 \times 2 \times 2 = 8;$ in the same way a set with $n$ elements has $2^{n}$ subsets.

Fig. 1.15Three yes/no choices give eight subsets. Solid branches mean "take", dashed branches mean "leave".

A tree that grows from left to right. In the first step a starting dot splits into two branches under the heading "take −2?": a solid branch for yes and a dashed branch for no. In the second step each of those dots splits again under "take 0?", giving four dots. In the third step each of the four splits under "take 2?", giving eight end dots labelled, from top to bottom, {−2, 0, 2}, {−2, 0}, {−2, 2}, {−2}, {0, 2}, {0}, {2} and empty braces. In the fourth step the count two times two times two equals eight appears below the tree.

Example 25NCERT Example 25 · p. 21

Equal differences force equal sets

Show that if $A - B = B - A,$ then $A = B$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

A \subset B

, which of these must be true?

B - A = ϕ

A \cap B = B

A \cup B = A

A - B = ϕ

Show answer

Answer (D) $A - B = ϕ$

Every element of

A

is in

B

, so nothing is left in

A - B

. The other three statements hold only when

A = B

Practice · 10 questions

Miscellaneous exercises on Chapter 1: questions with answers

Short answerMediumNCERT Miscellaneous Exercise, Q1
Let $A = {x : x \in 𝐑, x^{2} - 7 x + 10 = 0},$ $B = {2, 5, 8},$ $C$ the set of natural numbers that leave remainder 2 on division by 3, and $D = {5}$ . Which of these sets are subsets of which others?
Show solution
Answer $D \subset A \subset B \subset C,$ so also $D \subset B$ , $D \subset C$ and $A \subset C$ .
$x^{2} - 7 x + 10 = (x - 2) (x - 5),$ so $A = {2, 5}$ , and $C = {2, 5, 8, 11, \dots} .$
Then $D \subset A$ (5 is a root), $A \subset B$ , and $B \subset C$ because 2, 5 and 8 each leave remainder 2 on division by 3. Subsets of subsets are subsets, which gives the other three. No other inclusion holds: for example $8 \in B$ but $8 \notin A$ .
2
ProofHardNCERT Miscellaneous Exercise, Q2
Decide whether each statement is true or false. Prove the true ones and give a counterexample for the false ones.
1. Whenever $A \subset B$ , each $x \in A$ also satisfies $x \in B$ .
  TrueFalse
  Show solution
  True: every element of $A$ is an element of $B$ , and $x$ is an element of $A$ .
2. Whenever $B \in C$ and $A \in B$ , it follows that $A \in C$ .
  TrueFalse
  Show solution
  False. Take $A = {1}$ , $B = {{1}}$ and $C = {{{1}}}$ . Then $A \in B$ and $B \in C$ , but the only element of $C$ is ${{1}} \neq A$ , so $A \notin C$ .
3. Whenever $B \subset C$ and $A \subset̸ B$ , it follows that $A \subset̸ C$ .
  TrueFalse
  Show solution
  False. Take $A = {1, 3}$ , $B = {1}$ , $C = {1, 3}$ : $A \subset̸ B$ (3 is missing) and $B \subset C$ , yet $A \subset C$ .
4. Whenever $B \subset̸ C$ and $A \subset B$ , it follows that $A \subset̸ C$ .
  TrueFalse
  Show solution
  False. Take $A = {1}$ , $B = {1, 2}$ , $C = {1}$ : $A \subset B$ and $B \subset̸ C$ , yet $A \subset C$ .
ProofHardNCERT Miscellaneous Exercise, Q3
Let $A$ , $B$ and $C$ be sets with $A - B = A - C$ and $B - A = C - A .$ Show that $B = C$ .
Show solution
Let $x \in B$ . If $x \notin A$ , then $x \in B - A = C - A,$ so $x \in C$ .
If $x \in A$ , then $x \notin A - B$ (because $x \in B$ ). Since $A - B = A - C,$ also $x \notin A - C$ . As $x \in A$ , this forces $x \in C$ .
So $B \subset C$ in both cases. The hypotheses are symmetric in $B$ and $C$ , so $C \subset B$ as well, and $B = C$ .
ProofHardNCERT Miscellaneous Exercise, Q4
For subsets $A$ and $B$ of $U$ , show that these four conditions are equivalent: (i) $A \cap B = ϕ$ , (ii) $A \subset B^{'}$ , (iii) $A - B = A$ , (iv) $B \subset A^{'}$ .
Show solution
(i) ⇒ (ii): if $x \in A$ , then $x \notin B$ (otherwise $x \in A \cap B$ ), so $x \in B^{'}$ .
(ii) ⇒ (iii): $A - B \subset A$ always. If $x \in A$ , then $x \in B^{'}$ , so $x \notin B$ and $x \in A - B$ . Hence $A - B = A$ .
(iii) ⇒ (iv): let $x \in B$ . If $x$ were in $A$ , it would be in $A - B$ , so $x \notin B$ , a contradiction. So $x \in A^{'}$ .
(iv) ⇒ (i): if $x \in A \cap B$ , then $x \in B \subset A^{'},$ so $x \notin A$ , a contradiction. Hence $A \cap B = ϕ$ .
ProofMediumNCERT Miscellaneous Exercise, Q5
If $A \subset B$ , show that $B^{'} \subset A^{'}$ .
Show solution
Let $x \in B^{'}$ , so $x \notin B$ . If $x$ were in $A$ , then $x \in B$ because $A \subset B$ , which is false. So $x \notin A$ , that is, $x \in A^{'}$ . Hence $B^{'} \subset A^{'}$ .
6
ProofMediumNCERT Miscellaneous Exercise, Q6
Show that for any subsets $A$ and $B$ of $U$ , (a) $A - B = A \cap B^{'},$ and (b) $A - (A - B) = A \cap B .$
1. $A - B = A \cap B^{'}$
  Show solution
  $x \in A - B$ means $x \in A$ and $x \notin B$ , which means $x \in A$ and $x \in B^{'}$ , that is, $x \in A \cap B^{'} .$ Each step works in both directions, so the sets are equal.
2. $A - (A - B) = A \cap B$
  Show solution
  By (a), $A - (A - B) = A \cap (A \cap B^{'})^{'},$ and De Morgan's law turns $(A \cap B^{'})^{'}$ into $A^{'} \cup B$ .
  So the set is $A \cap (A^{'} \cup B) = (A \cap A^{'}) \cup (A \cap B) = ϕ \cup (A \cap B) = A \cap B .$
7
ProofMediumNCERT Miscellaneous Exercise, Q7
Using the laws of sets, show that:
1. $(A \cup B) \cap (A \cup B^{'}) = A$
  Show solution
  Union distributes over intersection, so the left side is $A \cup (B \cap B^{'}) = A \cup ϕ = A .$
2. $(A \cap B) \cup (A \cap B^{'}) = A$
  Show solution
  Intersection distributes over union, so the left side is $A \cap (B \cup B^{'}) = A \cap U = A .$
Short answerMediumNCERT Miscellaneous Exercise, Q8
Show by an example that $A \cup B = A \cup C$ does not force $B = C$ .
Show solution
AnswerFor example $A = {1, 2}$ , $B = {1}$ , $C = {2}$ .
With $A = {1, 2}$ , $B = {1}$ and $C = {2}$ , both unions equal ${1, 2}$ , but $B \neq C$ . The elements in which $B$ and $C$ differ are hidden inside $A$ .
ProofHardNCERT Miscellaneous Exercise, Q9
Let $A$ , $B$ and $X$ be subsets of $U$ with $A \cup X = B \cup X$ and $A \cup X^{'} = B \cup X^{'} .$ Show that $A = B$ .
Show solution
By the distributive law, $(A \cup X) \cap (A \cup X^{'}) = A \cup (X \cap X^{'}) = A \cup ϕ = A .$ In the same way $(B \cup X) \cap (B \cup X^{'}) = B .$
The two left-hand sides are equal term by term by the hypotheses, so $A = B$ .
Short answerMediumNCERT Miscellaneous Exercise, Q10
Find sets $A$ , $B$ and $C$ , each with three elements, such that any two of them have exactly one common element but $A \cap B \cap C = ϕ .$ How many elements does $A \cup B \cup C$ have?
Show solution
AnswerFor example $A = {1, 2, 3},$ $B = {3, 4, 5},$ $C = {5, 6, 1};$ the union has 6 elements.
$A \cap B = {3}$ , $B \cap C = {5}$ and $A \cap C = {1}$ , while no number lies in all three sets.
$A \cup B \cup C = {1, 2, 3, 4, 5, 6} :$ the nine places in the three sets hold six different numbers, since each of 1, 3 and 5 is counted twice.

Recap

Sets: key points and summary

A set is a well-defined collection, described in roster form or in set-builder form (Section 1.2).
The empty set $ϕ$ has no elements; ${0}$ and ${ϕ}$ are not empty (Section 1.3).
$n (A)$ counts distinct elements. A set is finite if it is empty or its count ends at a natural number; otherwise it is infinite (Section 1.4).
Equal sets have exactly the same elements; order and repetition never matter (Section 1.5).
$A \subset B$ when every element of $A$ is in $B$ . Always $ϕ \subset A$ and $A \subset A$ , and $A = B$ exactly when $A \subset B$ and $B \subset A$ (Section 1.6).
$\in$ links an element to a set, while $\subset$ links two sets (Example 11).
$𝐍 \subset 𝐙 \subset 𝐐 \subset 𝐑,$ and the irrationals $𝐓$ are the reals outside $𝐐$ (Section 1.6.1).
An interval with end points $a < b$ has length $b - a$ whether it is open, closed or half-open; a square bracket includes its end point (Section 1.6.2).
A universal set $U$ contains every set under discussion; a Venn diagram draws it as a rectangle with circles for its subsets (Section 1.8).
$A \cup B$ holds what is in either set and $A \cap B$ what is in both; disjoint sets have $A \cap B = ϕ$ . Each operation distributes over the other (Section 1.9.2).
$A - B$ keeps the elements of $A$ outside $B$ . Together, $A - B$ , $A \cap B$ and $B - A$ make up $A \cup B$ without overlapping (Section 1.9.3).
$A^{'} = U - A$ and $(A^{'})^{'} = A .$ By De Morgan's laws, the complement of $A \cup B$ is $A^{'} \cap B^{'}$ , and the complement of $A \cap B$ is $A^{'} \cup B^{'}$ (Section 1.10).

Flash cards · 8

Symbols at a glance

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From historyNCERT Historical Note · p. 22

How set theory began

Georg Cantor (1845–1918) came to sets around 1870 while asking when a function can be written as a sum of sines and cosines in only one way. The answer depended on a set of exceptional points, and he began to study such sets for their own sake. His most famous discovery came in 1874: the real numbers cannot be arranged in a list numbered 1, 2, 3, …, so infinite sets come in different sizes. The idea divided mathematicians. Leopold Kronecker in Berlin rejected proofs that claimed a set exists without constructing it, while Richard Dedekind (1831–1916) and, later, David Hilbert championed Cantor.

Around 1900 Gottlob Frege tried to build all of arithmetic on logic and sets. In 1902 Bertrand Russell (1872–1970) wrote to him with a paradox. Call a set ordinary if it is not an element of itself, and let $R$ be the set of all ordinary sets. If $R$ is ordinary, it belongs to $R$ , so it is not ordinary; if it is not ordinary, it does not belong to $R$ , so it is ordinary. Either way there is a contradiction: no set can contain everything.

The cure was to stop forming a set from any description whatsoever, and to list instead, as axioms, the ways in which new sets may be built from old ones. Ernst Zermelo proposed such axioms in 1908; refined by Abraham Fraenkel and others, they became the system known as ZF. The safe way of describing sets that you practise in this chapter, by picking elements out of a given universal set, is exactly what those axioms allow.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 1.1, Q1	Practice 1.1 · Q1
NCERT Exercise 1.1, Q2	Practice 1.1 · Q2
NCERT Exercise 1.1, Q3	Practice 1.1 · Q3
NCERT Exercise 1.1, Q4	Practice 1.1 · Q4
NCERT Exercise 1.1, Q5	Practice 1.1 · Q5
NCERT Exercise 1.1, Q6	Practice 1.1 · Q6
NCERT Exercise 1.2, Q1	Practice 1.2 · Q1
NCERT Exercise 1.2, Q2	Practice 1.2 · Q2
NCERT Exercise 1.2, Q3	Practice 1.2 · Q3
NCERT Exercise 1.2, Q4	Practice 1.2 · Q4
NCERT Exercise 1.2, Q5	Practice 1.2 · Q5
NCERT Exercise 1.2, Q6	Practice 1.2 · Q6
NCERT Exercise 1.3, Q1	Practice 1.3 · Q1
NCERT Exercise 1.3, Q2	Practice 1.3 · Q2
NCERT Exercise 1.3, Q3	Practice 1.3 · Q3
NCERT Exercise 1.3, Q4	Practice 1.3 · Q4
NCERT Exercise 1.3, Q5	Practice 1.3 · Q5
NCERT Exercise 1.3, Q6	Practice 1.3 · Q6
NCERT Exercise 1.3, Q7	Practice 1.3 · Q7
NCERT Exercise 1.3, Q8	Practice 1.3 · Q8
NCERT Exercise 1.4, Q1	Practice 1.4 · Q1
NCERT Exercise 1.4, Q2	Practice 1.4 · Q2
NCERT Exercise 1.4, Q3	Practice 1.4 · Q3
NCERT Exercise 1.4, Q4	Practice 1.4 · Q4
NCERT Exercise 1.4, Q5	Practice 1.4 · Q5
NCERT Exercise 1.4, Q6	Practice 1.4 · Q6
NCERT Exercise 1.4, Q7	Practice 1.4 · Q7
NCERT Exercise 1.4, Q8	Practice 1.4 · Q8
NCERT Exercise 1.4, Q9	Practice 1.4 · Q9
NCERT Exercise 1.4, Q10	Practice 1.4 · Q10
NCERT Exercise 1.4, Q11	Practice 1.4 · Q11
NCERT Exercise 1.4, Q12	Practice 1.4 · Q12
NCERT Exercise 1.5, Q1	Practice 1.5 · Q1
NCERT Exercise 1.5, Q2	Practice 1.5 · Q2
NCERT Exercise 1.5, Q3	Practice 1.5 · Q3
NCERT Exercise 1.5, Q4	Practice 1.5 · Q4
NCERT Exercise 1.5, Q5	Practice 1.5 · Q5
NCERT Exercise 1.5, Q6	Practice 1.5 · Q6
NCERT Exercise 1.5, Q7	Practice 1.5 · Q7
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises on Chapter 1 · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises on Chapter 1 · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises on Chapter 1 · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises on Chapter 1 · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises on Chapter 1 · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises on Chapter 1 · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises on Chapter 1 · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercises on Chapter 1 · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercises on Chapter 1 · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercises on Chapter 1 · Q10

Quick check: MCQs on Sets

1.2 Sets and their Representations4 questions
1.3 The Empty Set2 questions
1.4 Finite and Infinite Sets2 questions
1.5 Equal Sets2 questions
1.6.2 Intervals as subsets of R4 questions
1.8 Venn Diagrams2 questions
1.9.3 Difference of sets4 questions
1.10 Complement of a Set4 questions
Miscellaneous Examples2 questions

Glossary · 16

Key terms (English–Hindi)

complementपूरकpoorak: $A^{'}$ , the set of elements of the universal set $U$ that are not in $A$ ; so $A^{'} = U - A .$ First seen in the chapter →

difference of setsसमुच्चयों का अंतरsamuchchayon ka antar: $A - B$ , the set of elements of $A$ that are not in $B$ .
disjoint setsअसंयुक्त समुच्चयasanyukt samuchchay: Sets with no common element, so that $A \cap B = ϕ$ .First seen in the chapter →

empty setरिक्त समुच्चयrikt samuchchay: The set with no elements, written $ϕ$ or ${}$ .
equal setsसमान समुच्चयsamaan samuchchay: Sets with exactly the same elements; order and repetition in a listing do not matter.

finite setपरिमित समुच्चयparimit samuchchay: A set that is empty or whose elements can be counted with the count ending at a natural number; any other set is infinite.

intersectionसर्वनिष्ठsarvanishth: $A \cap B$ , the set of elements that belong to both $A$ and $B$ .First seen in the chapter →
intervalअंतरालantaraal: The set of all real numbers between two end points, each end point included or excluded, such as $[a, b)$ .

proper subsetउचित उपसमुच्चयuchit upasamuchchay: A subset $A$ of $B$ with $A \neq B$ , so that $B$ has at least one element outside $A$ .First seen in the chapter →

roster formरोस्टर रूपroster roop: A description of a set that lists its elements, separated by commas, inside braces, as in ${2, 4, 6}$ .First seen in the chapter →

setसमुच्चयsamuchchay: A well-defined collection of objects, called its elements: for any object it is certain whether it belongs.First seen in the chapter →
set-builder formसमुच्चय निर्माण रूपsamuchchay nirmaan roop: A description of a set by a property its elements share, as in ${x : x \in 𝐍, x < 4} .$ First seen in the chapter →
subsetउपसमुच्चयupasamuchchay: $A \subset B$ when each member of $A$ belongs to $B$ as well.

unionसम्मिलनsammilan: $A \cup B$ , the set of elements in $A$ or in $B$ or in both.First seen in the chapter →
universal setसार्वत्रिक समुच्चयsaarvatrik samuchchay: The set $U$ that contains every set under discussion in a given context.

Venn diagramवेन आरेखven aarekh: A picture of sets: a rectangle for the universal set and closed curves, usually circles, for its subsets.First seen in the chapter →

Sets: Class 11 Maths · Chapter 1

What you will learn in Sets

1.1 Introduction

1.2 Sets and their Representations

Practice 1.1: questions with answers

1.3 The Empty Set

1.4 Finite and Infinite Sets

1.5 Equal Sets

Practice 1.2: questions with answers

1.6 Subsets

1.6.1 Subsets of set of real numbers

1.6.2 Intervals as subsets of R

1.7 Universal Set

Practice 1.3: questions with answers

1.8 Venn Diagrams

1.9 Operations on Sets

1.9.1 Union of sets

1.9.2 Intersection of sets

1.9.3 Difference of sets

Practice 1.4: questions with answers

1.10 Complement of a Set

Practice 1.5: questions with answers

Miscellaneous Examples

Miscellaneous exercises on Chapter 1: questions with answers

Practice questions with answers (NCERT “Entrance-style practice”)

Sets: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Sets

Key terms (English–Hindi)

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